A uniform solid disk and a uniform hoop are placed side by side at the top of an incline of height h. If they are released from rest and roll without slipping, determine their speeds when they reach the bottom. (d = disk, h = hoop)

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sweetgoodnews sweetgoodnews · Answer 1 · 2020-02-27T05:32:36+01:00

Answer:

) the uniform disk has a lower moment of inertia and arrives first.

Explanation:

(a) the uniform disk has a lower moment of inertia and arrives first.

(b) Let's say the disk has mass m and radius r, and

the hoop has mass M and radius R.

disk: initial E = PE = mgh

I = ½mr², so KE = ½mv² + ½Iω² = ½mv² + ½(½mr²)(v/r)² = (3/4)mv² = mgh

m cancels, leaving v² = 4gh / 3

hoop: initial E = Mgh

I = MR², so KE = ½MV² + ½(MR²)(V/R)² = MV² = Mgh

M cancels, leaving V² = gh

Vdisk = √(4gh/3) > Vhoop = √(gh)

tallinn tallinn · Answer 2 · 2022-04-12T15:08:01+02:00

a. When the uniform disk has a lower moment of inertia and arrives first.

b. V disk is = √(4gh/3) > Vhoop = √(gh)

(a) When the uniform disk has a lower moment of inertia and arrives first.

(b) Now Let's say the disk has mass m and also radius r, and the hoop has mass M and radius R.

Then, disk: initial E = PE = mgh

After that, I = ½mr², so KE = ½mv² + ½Iω² = ½mv² + ½(½mr²)(v/r)² = (3/4)mv² = mgh

Now, m cancels, leaving v² = 4gh / 3

Then, hoop: initial E = Mgh

After that, I = MR², so KE = ½MV² + ½(MR²)(V/R)² = MV² = Mgh

Then, M cancels, leaving V² = gh

Therefore, V disk is = √(4gh/3) > Vhoop = √(gh)

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